Articles
Register
Sign In
Search
LaMissD
@LaMissD
January 2021
1
54
Report
Bonjour :)
...je suis désolé mais jai du mal avec cet exercice...pourrait on me donner un petit coupe pouce svp?
Merci d'avance
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
scoladan
Verified answer
Bonjour,
1)
f(2) = 20/ln(10) x ln(50000x2) = 20/ln(10) x ln(10^5) = 20/ln(10) x 5ln(10) = 100 dB
f(0,2) = 20/ln(10) x ln(50000x0,2) = 20/ln(10) x ln(10^4) = 20/ln(10) x 4ln(10) = 80 dB
f(0,02) = .... = 20/ln(10) x 3ln(10) = 60 dB
f(p0) = 20/ln(10) x ln(50000x20.10^-6)
= 20/ln(10) x ln(1) = 0 dB
2) f(p) = 120
<=> 20/ln(10) x ln(50000p) = 120
<=> ln(50000p) = 120xln(10)/20 = 6ln(10) = ln(10^6)
==> 50000p = 6
==> p = 6/50000 = 12/100000 = 12.10^-5 Pa
3) f(10x) = 20/ln(10) x ln(50000x10x)
= 20/ln(10) x ln(500000x) = 20/ln(10) + [ln(50000x) + ln(10)]
= 20/ln(10) x ln(10) + 20/ln(10) x ln(50000x)
= 20 + f(x)
donc f(10x) = 20 + f(x)
Quand la pression est multipliée par 10, le niveau sonore augmente de 20 dB.
4) f(100x) = f(10x10x) = 20 + f(10x) = 20 + 20 + f(x) = 40 + f(x)
Quand la pression est multipliée par 100, le niveau sonore augmente de 40 dB.
1 votes
Thanks 1
LaMissD
Merci beaucoup!!!
More Questions From This User
See All
LaMissD
January 2021 | 0 Respostas
Responda
LaMissD
January 2021 | 0 Respostas
Responda
LaMissD
January 2021 | 0 Respostas
Responda
LaMissD
January 2021 | 0 Respostas
Responda
LaMissD
January 2021 | 0 Respostas
Responda
Recomendar perguntas
mehdizone4
November 2022 | 0 Respostas
carla3096
November 2022 | 0 Respostas
lilastark77
October 2022 | 0 Respostas
louise5654
October 2022 | 0 Respostas
charliegumezy
October 2022 | 0 Respostas
maguyungudimayoyolod
October 2022 | 0 Respostas
nounou8448
October 2022 | 0 Respostas
kenan93
October 2022 | 0 Respostas
juju56789
October 2022 | 0 Respostas
nhvabesoindaidesvp
October 2022 | 0 Respostas
×
Report "Bonjour :) ...je suis désolé mais jai du mal avec cet exercice...pourrait on me donner un petit coup.... Pergunta de ideia de LaMissD"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
Verified answer
Bonjour,1)
f(2) = 20/ln(10) x ln(50000x2) = 20/ln(10) x ln(10^5) = 20/ln(10) x 5ln(10) = 100 dB
f(0,2) = 20/ln(10) x ln(50000x0,2) = 20/ln(10) x ln(10^4) = 20/ln(10) x 4ln(10) = 80 dB
f(0,02) = .... = 20/ln(10) x 3ln(10) = 60 dB
f(p0) = 20/ln(10) x ln(50000x20.10^-6)
= 20/ln(10) x ln(1) = 0 dB
2) f(p) = 120
<=> 20/ln(10) x ln(50000p) = 120
<=> ln(50000p) = 120xln(10)/20 = 6ln(10) = ln(10^6)
==> 50000p = 6
==> p = 6/50000 = 12/100000 = 12.10^-5 Pa
3) f(10x) = 20/ln(10) x ln(50000x10x)
= 20/ln(10) x ln(500000x) = 20/ln(10) + [ln(50000x) + ln(10)]
= 20/ln(10) x ln(10) + 20/ln(10) x ln(50000x)
= 20 + f(x)
donc f(10x) = 20 + f(x)
Quand la pression est multipliée par 10, le niveau sonore augmente de 20 dB.
4) f(100x) = f(10x10x) = 20 + f(10x) = 20 + 20 + f(x) = 40 + f(x)
Quand la pression est multipliée par 100, le niveau sonore augmente de 40 dB.